If $s \in L(r') \cup L(r'')$, then it is clear that $s$ has the form $s = \omega_1 b \omega_2 a \omega_3$, where $\omega_k \in \Sigma^*$, and so $s \in L(r)$. Hence $L(r') \cup L(r'') \subset L(r)$.

Now suppose $s \in L(r)$. From the above reflection, we see that $s$ has the form $s=\omega_1b b^* a \omega_2$, with $\omega_k \in \Sigma^*$. Hence $ s \in L(r')$. Now let $i$ be the first index such that $s_i = b$ and let $j$ be the last index such that $s_j = a$. (Since $s \in L(r)$ we know that such indices exist and that $i < j$.) Then $s$ has the form $a^* b \omega_1 a b^*$, with $\omega_1 \in \Sigma^*$. Hence it follows that $s \in L(r'')$.

I am not sure what you want. I have shown (1) $L(r') \cup L(r'') \subset L(r)$, (2) $L(r) \subset L(r')$ and (3) $L(r) \subset L(r'')$. That pretty much covers it. You can separate the proof out if you want, but all the details are there...
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copper.hatNov 22 '12 at 20:15